Question

differentiation of Tan[Sec²x]​

Answer 1

$\large\underline{\sf{Solution-}}$

Given function is

$\rm :\longmapsto\:y = tan( {sec}^{2}x)$

On differentiating both sides w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{d}{dx}y =\dfrac{d}{dx} tan( {sec}^{2}x)$

We know, .

$\rm :\longmapsto\:\boxed{\tt{ \dfrac{d}{dx}tanx = {sec}^{2}x}}$

So, using this, we get

$\rm :\longmapsto\:\dfrac{dy}{dx} = {sec}^{2}( {sec}^{2}x)\dfrac{d}{dx} {sec}^{2}x$

can be rewritten as

$\rm :\longmapsto\:\dfrac{dy}{dx} = {sec}^{2}( {sec}^{2}x)\dfrac{d}{dx} {(secx)}^{2}$

We know,

$\rm :\longmapsto\:\boxed{\tt{ \dfrac{d}{dx} {x}^{n} = {nx}^{n - 1}}}$

So, using this, we get

$\rm :\longmapsto\:\dfrac{dy}{dx} = {sec}^{2}( {sec}^{2}x) \times 2(secx)\dfrac{d}{dx} {(secx)}$

$\rm :\longmapsto\:\dfrac{dy}{dx} = 2secx \: {sec}^{2}( {sec}^{2}x) (secx \: tanx)$

$\rm\implies \:\boxed{\tt{ \dfrac{dy}{dx} = 2 \: tanx \: {sec}^{2}x \: {sec}^{2}( {sec}^{2}x) \: }}$

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Additional Information

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}$